2016 -2017 Soil Third - الجامعة التكنولوجية€¦ · Mohr circle: it is concerned...
Transcript of 2016 -2017 Soil Third - الجامعة التكنولوجية€¦ · Mohr circle: it is concerned...
Soil
Mechanics
Lectures
2016-2017
Third Year
Students Includes: Stresses within the soil, consolidation theory, settlement and
degree of consolidation, shear strength of soil, earth pressure on retaining
structure.:
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Stresses within the soil
Stresses within the soil:
Types of stresses:
1- Geostatic stress: Sub Surface Stresses cause by mass of soil
a- Vertical stress �� = ∑�ℎ
b- Horizontal Stress �� =��
Note : Geostatic stresses increased lineraly with depth.
2- Stresses due to surface loading :
a- Infintly loaded area (filling)
b- Point load(concentrated load)
c- Circular loaded area.
d- Rectangular loaded area.
Introduction:
At a point within a soil mass, stresses will be developed as a result of the soil
lying above the point (Geostatic stress) and by any structure or other loading
imposed into that soil mass.
1-
stresses due Geostatic soil mass
�� = �ℎ (Geostatic stress)
�� =��� , where � : is the coefficient of earth pressure at
rest.
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EFFECTIVESTRESS CONCEPT: In saturated soils, the normal stress (σ) at any
point within the soil mass is shared by the soil grains and the water held within the
pores. The component of the normal stress acting on the soil grains, is called
effective stressor intergranular stress, and is generally denoted by σ'. The
remainder, the normal stress acting on the pore water, is knows as pore water
pressure or neutral stress, and is denoted by u. Thus, the total stress at any point
within the soil mass can be written as: � = ��+ u
This applies to normal stresses in all directions at any point within the soil mass. In
a dry soil, there is no pore water pressure and the total stress is the same as
effective stress. Water cannot carry any shear stress, and therefore the shear stress
in a soil element is carried by the soil grains only.
1- W.T.L very deep
�� = � �����ℎ�����.����������
2- When W.T.L. at the ground surface:
�� = � �����ℎ�����.����������
u= h*�� �� =�� − = ��!" = ���� = ���� −��
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2- Stresses due to filling: ∆����� =ℎ���� ∗ ����� 3-Stresses due to external loading:
a- Stresses due to point load :
Boussinesq (1885)(French mathematician) developed the following
equation to calculate vertical stress increment ∆�, in soils due to point
load on the surface:
Type of point load: the load under
the wheel of car.
Then, ∆� = %&' ()
Where factor, (*
(* = 32- . 11 + 123456
75
Example 1:
In road pavement design, the standard vehicle axel is defined as an axel with two
single wheels as shown below. For a particular vehicle group, the standard axel load
(P) is given as 80 kN and the distance between two wheels (L) is 1.8 m. what is the
vertical stress increment in the sub-grade at 4 m depth directly under a wheel if this
axel is running (consider wheel load as a point load ):
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Solution:
∆� = %&' (),
(* = 32- . 11 + (23)567/5
To find ∆� = ∆�; +∆�<
((=); = 32- . 11 + (23)567/5
= 32-. 11 + (1.84 )56
7/5= 0.301
((=)< = 32- . 11 + (23)567/5
= 32-. 11 + (04)56
7/5= 0.477
Therefore,
∆� = B<C5 ((=)< +B;C5 ((=); =4045 ∗ 0.477 +4045 ∗ 0.301 = 1.945�FG
b- Circular footing ∆�� = H 1IJ , 3J4 ∗ ∆L�MNGO
F: factor, can be find from the following figure PQ , where R: is the radius of the circular area
X: is the distance from the center of the circle to the point
Z : is the depth of the point.
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c- Rectangular loaded area. ∆��R���������� = S(M,T) ∗ L����� F(n,m) : is the factor (use figure below)
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Approximate method to find ∆� at Z using 2:1 method:
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∆� = L����∗;∗U(V + 3) ∗ (W + C) WhereL����: Is the net surface load.
Example: Plot the variation of total and effective vertical stresses, and pore water
pressure with depth for the soil profile shown below:
Solution:
Within a soil layer, the unit weight is constant, and therefore the stresses vary
linearly. Therefore, it is adequate if we compute the values at the layer
interfaces and water table location, and join them by straight lines. At the
ground level,
σv= 0 ; �� = 0; and u=0 At 4 m depth,
σv= (4)(17.8) = 71.2 kPa; u = 0
∴�� = 71.2 kPa
At 6 m depth, σv= (4)(17.8) + (2)(18.5) = 108.2 kPa u = (2)(9.81) = 19.6 kPa
∴ �� = 108.2 – 19.6 = 88.6 kPa At 10 m depth, σv= (4)(17.8) + (2)(18.5) + (4)(19.5) = 186.2 kPa u = (6)(9.81) = 58.9 kPa
∴�� = 186.2 – 58.9 = 127.3 kPa
At 15 m depth,
σv= (4)(17.8) + (2)(18.5) + (4)(19.5) + (5)(19.0) = 281.2 kPa u = (11)(9.81) = 107.9 kPa
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∴��= 281.2 – 107.9 = 173.3 kPa The values of σv, u and
�� Computed above are summarized in Table 6.1.
Variation of ��, GMY�� with depth Negative pore pressure (suction):
Below the water table, pore pressure are positive. In dry soil, the pore
pressure are positive. In dry soil, the pore pressure is zero. Above the
water table, when the soil is saturated, pore pressure will be negative.
u=−ℎ���
The height above the water table to which the soil is saturated is called
the capillary rise, and this depends on the grain size and type(and thus
the size of pores):
-In coarse soils capillary rise is very small
-In silts it may be up to 2m
_In clays it can be over 20m
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Example :
For the soil profile shown find the total, effective and pore water pressure:?
Given: ���� = 16�[T\ , ���� = 20 �[T\
Solution:
At depth
z=0, ������ = 0, = 0, ���� = 0
At depth
z= 2m ������ = 2 ∗ 16 = 32 ]_' , = 0, ���� = 32 ]_' At depth
Z= 5 m ������ = 2 ∗ 16 + 3 ∗ 20 = 92 ]_' , = 3 ∗ 10 = 30 ]_' , ���� = 32 ]_'
0
1
2
3
4
5
6
0 20 40 60 80 100
Total Stress
Pore water
pressure
Effective stress
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Example:
For the soil profile shown: Determine , total, effective and pore water pressure for the following
conditions:1- water table 3 m above the ground level, 2- water table at the ground level 3- 1.0 m
below the surface,4- 2 m below the surface?
���� = 16 �[T\, ���� = 20 �[T\
Depth
(m)
WT Total stress(kPa) u(kPa) ���� (�FG) 0 Case 1 30 30 0
3 30+3*20=90 6*10=60 30
0 Case 2 0 0 0
3 3*20=60 3*10=30 30
0 Case 3 0 0
1 1*16=16 0 16
3 16+20*2=56 20 36
0 Case 4 0 0 0
2 2*16=32 0 32
3 32+1*20=52 1*10=10 42
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40 50
Case
1/effectivecase 2effective
case
3/effective
Stresses (kPa)
de
pth
(m
)
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Example :
A building 20 m * 20m results in a uniform surface contact pressure of 150 kPa.
Determine the increase in vertical pressure at depth of 10 m below a) the center of
the building b) the corner of the building. Estimate the additional pressure at both
locations of a tower 5m *5m placed at the center of the building imposing 300 kPa
uniform additional pressures.
At the corner:
nz= 10 n*10=20 , n=2
mz=10 m*10=20, m=2
F(1,1)= 0.2325 ∆�GOOℎN`a2MN2Y NOab 2HG`NcaGY =0.2325 ∗ 150 = 34.875�FG
At the center :
N*10=10
n=m=1
From figure (rectangular ) f(1, 1)= 0.085
∆� at the center =4*0.176*150= 105.6 kPa ∆�Y NOaOadN2 =
at the center:
n*10=2.5
n=m= 0.25
f(0.25,0.25)= 0.0261
∆�Y NOaOadN2eNcadOℎN`NMON2 ∆� = 4*0.0261*300=31.32 kPa
∆� at the corner due to tower
for e1:
n*10=12.5
n=m= 1.25
f(1.25,1.25)=
0.211
For e2=e3
n*10=7.5
n=0.75
m*10=12.5
m=1.25
f(0.75,1.25)=
0.165
for e3:
n*10=7.5,
n=m=0.75
f(0.75,0.75)=0.136
=(0.211-(0.165*2)-0.136)*300=5.1
kPa
Top Veiew
Front Veiew
20 m * 20m
5 m
5 m
5 m
12.5 m
12.5 m
12.5 m
12.5 m +e1
-e2
7.5 m
12.5 m
+e3
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Example: A distributed load of 50 kN/T5 is acting on the flexible rectangular area
6*3 m, as shown in figure. Determine the vertical stress at point A which is located
at depth of 3 m below the ground surface. (� = 18.5�[/T5).
Solution:
e1=e3, n*3=4.5 -------n=1.5, m*3= 1.5, m= 0.5
f(1.5,0.5)=0.131
e2=e4 , n*3= 1.5 , n=0.5, m*3=1.5,m=0.5
f(0.5,0.5)=0.085
∆� = 50 ∗ (0.131 ∗ 2 + 2 ∗ 0.085) = 21.6�[/T5
Total stresses = Geostatic stress+∆�
Total Stresses= 18*3+21.6=75.6 kPa
Example : For the same example .If the foundation at 0.5 m below ground surface:
q net= 50- 0.5*18= 41 kN/T5
e1=e3, n*2.5=4.5, n=1.8, m*2.5= 1.5, m=0.6
f ( 1.8,0.6) =0.155
e2=e4, n*2.5=1.5, n= 0.6, m*2.5=1.5, m=0.6
f (0.6,0.6)= 0.109
∆� = 41 ∗ (0.155 ∗ 2 + 2 ∗ 0.109) = 21.648�[/T5
Total stresses= 18*2.5+21.648=66.648 kN/T5
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Principal Stresses and Mohr Circle
Principal Stresses:
There are exist at any stressed point in three orthogonal (i.e. naturally
perpendicular) planes on which there are zero shear stresses. These planes are
called the principal stress planes. The normal stresses that acts on these three planes
are principal stresses. The largest of these three stresses is called the major
principal stress �� , the smallest is called the minor principal stress�\, and the third
is called the intermediate stress�5. When the stresses in the ground are geostatic,
the horizontal plane through a point is a principal plane and so too are all vertical
planes through the point. When ( � = fgfh) � < 1, ��= �� and ��= �\ = �5 .
when � > 1, the situation is reversed. The shear stresses on any orthogonal
planes (planes meeting at right angles) must be numerically equal (k� =k�). Mohr circle: it is concerned only with the stresses existing in two dimensions, the
state of stress in plane that contains the major and minor principal stresses �� ,�\.
The stresses will be considered positive when compressive.
Equations for state of stress at a point. k is positive when counter clock wise l is measured counter clockwise from the
direction of ��.
Mohr diagram for state of stress at a point
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�m =�� +�\2 +�� − �\2 cos 2l
k = (�� −�\)2 sin 2l
Where �m and k are the stresses acts on any planes, the direction and magnitude
of principal stresses can be found.
Origin of planes: is a point on the Mohr circle, denoted by Op, with following
property:
1- A line through Op and any points (A) of the Mohr circle will be parallel to
the plane on which the stresses given by point A act.
2- If the plane and Op are known then the line parallel to the plane passes
through Op and intersect Mohr circle , the intersect point represent the point
which acts on that plane.
To find the stresses , there are two method:
1- By drawing (prefer)
2- By using equations
1- By drawing
To determine the normal and shear stresses on any plane , we have to do
the following:
1-Draw the stress- state on Mohr circle (compression (+) and ( ).
2-Find the point denoted by Op origin of planes by either select �� and draws a line
parallel to the plane on which ��is acting until it intersects Mohr Circle & the
point of intersection is OP, or use �\ (the same principal is apply to �\) .
3-From OP draw a line parallel to the plane you want to find stress on it. The point
of intersection with Mohr Circle represents the stress & shear stress you need.
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: ملخص
:طريقة الرسم
تبطة مع بعضھا ھي عوامل رئيسية مر) 3( ھناك
stressا&جھاد -1
Planeمستوي -2
OPنقطة اصل المستوي -3
OP :ھي نقطة وھمية وحيدة لكل دائرة ووحيدة فقط وتعين من اجھاد ومستوي.
.ويقطع دائرة مور فان ھذا الخط يمثل المستوي الذي تعمل عليه ذالك ا&جھاد OPاي خط يمر في
Principle Stress:
Stresses acting normal on mutually orthogonal planes with no shear stresses.
Principle Planes:
The planes on which there is zero shear stresses.
Case one: Given stuvwsx required syuvwzy
Example 1:
Find stresses on plane B-B?
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Answer:
�� = 40�[/T5 �\ = 20�[/T5
{ = ��|�\2 = 40 + 202 = 30�[/T5
J = f}~f�5 = �~55 = 10�[/T5
By drawing:
1- Locate points (40,0) and (20,0).
2- Draw circle, using these points to define diameter.
3- Draw line ���� through point (20,0) and parallel to plane on which stress
(20,0)acts.
4- Intersection of ���� with Mohr circle at point (40,0) is the origin
Of planes.
5- Draw Line V� V� through OP and parallel to B-B
6- Read coordinates of X where V� V� intersect Mohr circle.
So on B-B �� = 25)b�k = −8.7)b��
Method 2 using Equations :
�m =�� +�\2 +�� − �\2 cos 2l
�m = �|55 +�~55 cos( 2 ∗ 120) = 30 + (−5) = 25
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k = (�� −�\)2 sin 2l
k = (40 − 20)2 sin(2 ∗ 120) = −8.66�[/T5
Case 2 given syuvwzy , required stuvwsx
Using Equation:
��m =∑�� +�\2
�� − �\2 = �(�m −�� + �\2 )5 + km5 We usually take the largest couple.
�� =�� + �\2 +�� − �\2
�\ =�� + �\2 −�� − �\2
b�M2l = km�� − �\2 = 2km�� − �\
Example 2 : Find the stresses on Horizontal plane D-D?
1-Draw point 1(40,0), point (20,0)
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2- From point (20,0) draw a line through point (20,0) ,parallel to the plane
that the force (20,0) acts .
3- The point of intersection with Mohr –circle represent the OP.
4- From the op draw a horizontal line (line parallel to horizontal plane) the
point of intersection with Mohr circle represent the stresses on horizontal
plane. ( 35, 8.7).
Example 3 :
Find the Magnitude and direction of priciple stresses.
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1- Locate points (40,-10) and (20,10)
2- Erect diameter and draw Mohr circle.
3- Draw V� V� through (40,-10) parallel to BB
4- Intersection of V� V� with circle gives Op.
5- Read ��GMY�\ from graph.
6- Line through Op and �� give plane on which �� acts.
Solution by Equations :
1- Make use of that the sum of normal stresses is a constant: �� + �\2 = ∑ �m2 = 40 + 202 = 30)b� 2- Use the following equation:
1�� −�\2 4 = ���m − (�� + �\2 )�5 +�km�5
(f}~f�5 ) = ��20 − 30�5 + �10�5 =√200 = 14.14)b� 3- �� =1f}|f�5 4 + 1f}~f�5 4 = 44.14)b�
�\ =1f}|f�5 4 − 1f}~f�5 4 = 15.86)b� 4- Use stress pair in which �m�bcG2�NbO; (40,−10) b�M2l = 2km�� − �\ = −2028.28 = −0.707
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2l = −45° l = −22.5° 5- Angle from horizontal to major principle stress direction = 52.5°
Examples : to be added
Stress Path (p-q diagram):
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Chapter Seven
Consolidation of soil
A stress increase caused by the construction of foundations or other loads
compresses the soil layers. The compression is caused by (a) deformation of soil
particles, (b) relocations of soil particles, and (c) expulsion of water or air from the
void spaces. In general, the soil settlement caused by load may be divided into
three broad categories:
1. Immediate settlement, which is caused by the elastic deformation of dry soil and
of moist and saturated soils without any change in the moisture content. Immediate
settlement calculations are generally based on equations derived from the theory of
elasticity.
2. Primary consolidation settlement, which is the result of a volume change in
saturated cohesive soils because of the expulsion of water that occupies the void
spaces.
3. Secondary consolidation settlement, which is observed in saturated cohesive
soils and is the result of the plastic adjustment of soil fabrics. It follows the primary
consolidation settlement under a constant effective stress.
This chapter presents the fundamental principles for estimating the consolidation
settlement:
Consolidation : is the gradual reduction in volume of a fully saturated
soil of low permeability due to drainage of some of the pore water, the
process continuing until the excess pore water pressure set up by an
increase in total stress has completely dissipated.
The simplest case is that of one – dimensional consolidation in which
condition of zero lateral strain.
Swelling: is the reverse of consolidation, is the gradual increase in
volume of a soil under negative excess pore water pressure.
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Consolidation settlement: is the vertical displacement of the surface
corresponding to the volume change at any stage of the consolidation
process.
Consolidation settlement will result, for example, if a structure is built
over a layer of saturated clay or if the water table is lowered permanently
in a stratum overlying a clay layer on the other hand, if an excavation is
made in a saturated clay, heaving (the reverse of settlement) will result in
the bottom of the excavation due to swelling of the clay.
Fundamentals of Consolidation: When a saturated clayey soil layer is
subjected to a stress increase, the pore water pressure suddenly increases.
So at time =0 ∆ = ∆� while ∆�=0 (Figure 1-b), but after a time t ∆ < ∆� while ∆� > 0 (Figure 1-c) and after very long time (O ≈∞)∆ = 0 while∆� = ∆� (Figure 1-d).
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.
Figure
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Spring Analogy: the following figures illustrate the soil-spring analogy:
Figure (7-2)
The Oedometer (consolidation)Test :
Figure (7-2) Consolidation Cell
Test Procedure:
1- Determine All data for soil sample such as diameter, weight and height. Also
determine Gs, initial water content , initial void ratio by using S*e= Gs .
W(s=100%).
2- Set the sample in the consolidation test and apply initial stress p1= 25 kPa
and record the dial reading for a period of (24 hrs) at the times 0,0.25, 0.5,
1,2,4,8,15, 30, 60, 120,240,480,1440 min. from the load application.
3- At the end of (24 hrs) , double the applied stress (p2= 2p1=50 kPa
Dial gage
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And record the dial gage reading for 24 hrs at similar times to those of step
2.
4- Repeat step 3 by doubling the applied stress and recording the dial gage
reading. This process is repeated till we reach a stress of 1600 kPa (some
times we reach 3200 kPa). This process last 7 days and it called loading
stage.
5- Unload (3/4) the applied stress (i.e. remove 1200 kPa and leave 400 kPa) and
record the dial gage reading for (24 hrs).
6- After 24 hrs , unload (3/4) the remaining stress (remove 300 kPa and keep
100 kPa)and record dial reading for 24 hrs. then remove all the applied stress
and record dial readings.
7- Determine the final water content of the soil sample.
Example : The results of an odometer test is given below:-
Time
(min)
Dial gage readings
25
kPa
50
kPa
100
kPa
200
kPa
400
kPa
800
kPa
1600
kPa
400
kPa
100
kPa
0
0 0 127.5 218 320 443 598 841 1038 1028 597
0.25 95 164 250 366 510 695 904 1038 937
0.5 101 168 255 373 515 704 910 1038 924
1 104 172 262 381 528 725 920 1036 904
2 110 179 268 391 544 739 932 1035 880
4 115 185 278 402 560 763 946 1033 835
8 115 189 284 413 570 783 963 1031 800
15 121 195 291 421 582 802 977 1030 765
30 124 198 300 428 591 805 979 1029 680
60 125.5 204 303 430 596 822 1000 1028 655
120 126 205 306 435 597 831 1012 1028 622
1440 127.5 218 320 443 598 841 1038 1028 597 420
Mass of empty ring = 99.18 gm
Mass or ring +wet soil = 266.4 gm
Mass of ring + dry soil = 226.68 gm
Dia. Of ring = 75 mm
Ht of ring = 19 mm
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Dial gage Coeff. = 0.0001"
Initial water content = 31.14%
Specific Gravity of soil solids (Gs) = 2.76
Initial Degree of saturation = 100%
Required: void ratio at each load increment.
Solution :
1- find N using S.e = Gs * �
N = 2.76 ∗ 0.3114
N = 0.8595
2- Since the volume of solid particles not change (incompressible)
So Assume �� = 1. So N = ��
So the Total volume (�� = 1 + N) So any change in the volume due to apply load is due to change in void so
∆N = N −N�
∈�= ∆N1 + N
Since the consolidation in one dimension, and there is no lateral strain so the axial
strain (∈�= ∆�� ) will be equal to volumetric strain (∈�). ∆�� = ∆��|��
Voids
Solids
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Applied
Pressure
(kPa)
Final
Dial gage
reading
Dial change
*0.0001*25.4
mm ∆ℎ(TT) Thickness
of sample
at the end
of (24 hrs)
mm
Change in
void ratio ∆N= ∆ℎ� (1+ N)
Void
ratio at
the end of
24 hrs
e=N −∆�
0
25
50
100
200
400
800
1600
400
100
0
0
127.5
218
320
443
598
841
1038
1028
597
420
0.32385
0.22987
0.25908
0.31242
0.3937
0.61722
0.5
0.0254
-1.09474
-0.4496
19
18.676
18.446
18.187
17.875
17.481
16.864
16.364
16.389
17.484
17.934
0.0317
0.0225
0.0254
0.0306
0.0385
0.0604
0.0489
-0.0025
-0.10714
-0.044
0.8595
0.8278
0.8053
0.7799
0.7493
0.07108
0.6504
0.6015
0.604
0.7111
0.755
From consolidation test the following result we can get :
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F�
The slop of this curve at any stress
range can be defined as : G� = ∆N∆F� G� : Coeff. Of compressibility ∆F ∶� change in effective stress ∆N ∶corresponding change in void
ratio
Figure e- F� (results from oedometer test)
LogF�
T� = G�1 + N T�: Coeff. Of volume change G� : Coeff. Of Compressibility N: Initial void ratio
Plot the Wa�F� �N2b b�a�Y2GO�a(N)* the slop of first portion of the
loading curve is defined {� = ∆N∆ca�F� Which is equal to the slope of
the unloading curve, and the
first portion of the reloading
curve ∴ {� ={� = ∆N∆ca�F� {�: Reloading index
Compression
Recompression or
Reloading
Unloading
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{�: Expansion index
The slope of the last portion of the loading curves is defined as :
{� = ∆N∆ca�F� {�: Compression index
Note : ∗ {� can be obtained using empirical correlation from Liquid Limit (L.L)
{� = 0.009(W. W − 10) An empirical correlation between {�GMY {� is given by:
{� ≅ 0.1{�
LogF�
Figure e- log ) (find F�� )
Pre-Consolidation
Pressure (F�� ) : The max. Effective stress
that has been experienced
by the soil in the past or at
the present.
How to find F��
1- Produce back the
straight line (BC) of
the curve
2- Draw the tangent to
the curve at D and
bisect the angle
between the tangent
and the horizontal
through D
3- The vertical through
the point of the
bisector and CB
gives the
approximate value
of the pre
consolidation
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pressure.
Normally and Over consolidation Clays:
A cording to the stress history, the clay can be:
1- Normally Consolidated clay(N.C.C.)
Which represent the clay at which the existing effective overburden pressure
(F� = ��) is the largest stress experienced by the soil at the present time
and in the past thus :
For N.C.C. F�� = F� 2- Over Consolidated Clay(O.C.C.)
Represent the clay which has experienced a stress in the past larger than the
existing effective overburden pressure (F� ) acting at the present time.
For O.C.C. F�� > F� Over Consolidation Ratio: (O.C.R.)
O.C.R.= =��=��
If O.C.R. =1 ∴ F�� = F� ∴ [. {. {. If O.C.R. >1 ∴ F�� > F� ∴ �. {. {. Consolidation Settlement:
To calculate the final consolidation settlement ( at t= ∞) use one of the following
:
1- �`H = ∆��|�� �
2- �`H = T�∆��
3- �`H = ���|�� � log =�� |∆=�=��
4- For O.C.C. we have to Check:
a- If F� + ∆F� ≤ F�� then use :
�`H = �¡�|�� � log =�� |∆=�=��
b- If F� + ∆F� > F�� then use :
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�`H = {�1 + N � log F��F� + {�1 + N � logF� + ∆F�F��
Where Scf : Final consolidation settlement
N: Initial void ratio
∆N: Change in void ratio = N −N�
H : thickness of compressed clay layer
T� : Coefficient of volume compressibility for the stress range
∆F� : change in effective stress between initial and finial conditions
{� ∶Compression index
{�: reloading index
F� : initial effective overburden pressure.
F�� : Pre consolidation pressure
Note About settlement Calculation:
1- Scf : usually find at the middle of compressed clay layer (represent the worse
case).
2- Calculate the increase in vertical stress at the middle of clay layer by either
using 2:1 method or using Chart (figures 8.6 and 8.4).
3- Check whether the clay is N.C.C. or O.C.C., and use proper equation
mention above.
4- For more accurate result you can divide the clay layer into sub layers (i.e not
more than 2 m). Then find the total settlement by summing the settlement for
each layer.
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Example:
For the soil Profile shown find the final consolidation settlement?
1- �`H = ∆��|�� � = ��~�¢�|�� � = �.£\~�.��|�.£\ (11.6 − 7.3) = 0.653T
2- �`H=T�∆F� �---------∆F� = ℎ���� ∗ ����� = 22 ∗ 4.5 = 99�[/T5
3- �`H = ���|�� � log =�� |∆=�=�� F� = � = � − at the middle of the clay layer before fill application
∴ � = (7.3 − 2) ∗ 18.22 ∗ ¤11.6 − 7.32 ¥ ∗ 16.34 = 131.7 �[T5
U= ¦(7.3 − 2.9) +(��.§~¨.\5 )© ∗ 9.807 = 64.236 ]_' F� = �131.7 − 64.236 = 67.46 ]_' �`H = �.ª77�|�.£\ (11.6 − 7.3) log §¨.�§|ªª§¨.�§ = 0.653T
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Degree of Consolidation (U% )
Degree of consolidation can be finding by
Figure: Assumed linear e-� relation
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Terzaghi's theory of one-dimensional consolidation:
The assumptions made in the theory are:
1- The soil is homogenous.
2- The soil is fully saturated.
3- The solid particles and water are incompressible.
4- Compression and flow is one-dimensional (vertical).
5- Strains are small.
6- Darcy's law is valid at all hydraulic GRADIENTS.
7- The coefficient of permeability and the coefficient of volume compressibility
remain constant throughout the process.
8- There is a unique relationship, independent of time, between void ratio and
effective stress.
There are three variable in the consolidation equation:
«!«� = {� «'!«&' (Consolidation equation)
1- Depth of the soil element in the layer (z).
2- The excess pore water pressure (U)
3- The time elapsed since application of the loading (t)
Where u: is the excess pore water pressure. {�: is the coefficient of consolidation.
{� = T���
Where T� ∶ is the coefficient of volume changes (T5 ¬NG2⁄ ).
®� ={�Oℎ��5
Where ®� ∶ Time factor
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t : the time ℎ��: is the drainage path
ℎ��=H/2 ( clay layer Between two permeable layers).
ℎ��= H (for clay layer between one permeable and one impermeable layer).
¯&: Degree of Consolidation
There is a relation between the ®� and Degree of consolidation
1- Degree of consolidation at specific depth
So the graphical solution of Terzaghi's one-dimensional consolidation
equation using non-dimensional parameters is as shown in following figure:
Figure Variation of degree of consolidation with time factor and
C ℎ��⁄ .
ℎ��= H
ℎ�� ℎ��= H
Cℎ��
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The variation of total consolidation with time is most conveniently plotted in the
form of the average degree of consolidation ( U ) for the entire stratum versus
dimensionless time ®�, and the this is illustrated below:
Figure : Average Degree of consolidation and Time factor
There are useful approximation relating the degree of consolidation and the time
factor
For ¯ < 0.60®� = °� ¯5
For ¯ > 0.60®� = 1.781 − 0.933ca�10(100- U)
Determination of coefficient of consolidation:
There are two methods:
1- The log time method (casagrande method):
a)-By plotting the dial reading versus log time
Then choice any time (small time ) then find the dial reading at 4t then move
above the dial reading a distance equal to ∆O then you can find the dial
reading at t= 0.
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b)-Take the tangent to the curves at the end of the curve so you can find the
dial reading at 100% consolidation.
c)-Now dial reading at t= 09 and dial reading at the end of consolidation so
you can find the dial reading for 50% consolidation.
e)-So the coefficient of consolidation `� can be find by the following
equation.
`� =0.196O5O7
2- The root time method (Taylor)
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{�= .£�£�±¡'�²�
Consolidation Settlement Time:
To estimate the amount of consolidation which occur and the time, it is necessary
to know:
1- The boundary drainage conditions
2- The loading condition
3- The soil parameters including initial void ratio, coefficient of
compressibility, compressibility index and coefficient of consolidation. All
these are obtained from consolidation test on representative sample.
So to plot the variation of settlement versus time curve:
1- Find the time required to complete the consolidation ( ®� = 1). 2- Assume different value of time start from (t=0) to time (t) find in step 1
3- For each value of time t , find the ®� using ( ®� = �g��±¡' ). 4- For each value of ®� (from step 3) find ( ¯��). 5- Find the final consolidation ( ���)using any proper equations.
6- Find the settlement at time t ³��� =¯�� ∗ ���´. 7- Plot the settlement-Time curve.
Time (t) ®� ={�Oℎ��5 ¯�� ��� ��� =¯�� ∗ ���
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Figure Variation of settlement with time
Example: For the soil Profile shown find the final consolidation settlement?
1- �`H = ∆��|�� � = ��~�¢�|�� � = �.£\~�.��|�.£\ (11.6 − 7.3) = 0.653T
2- �`H=T�∆F� �---------∆F� = ℎ���� ∗ ����� = 22 ∗ 4.5 = 99�[/T5
3- �`H = ���|�� � log =�� |∆=�=��
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F� = � = � − at the middle of the clay layer before fill application
∴ � = (7.3 − 2) ∗ 18.22 ∗ ¤11.6 − 7.32 ¥ ∗ 16.34 = 131.7 �[T5
U= ¦(7.3 − 2.9) +(��.§~¨.\5 )© ∗ 9.807 = 64.236 ]_' F� = �131.7 − 64.236 = 67.46 ]_' �`H = �.ª77�|�.£\ (11.6 − 7.3) log §¨.�§|ªª§¨.�§ = 0.653T
Example :
Astrata of consolidated clay of thickness 10 ft drained on one side only.
The hydraluic conductivity of k = 1.863*10~£ in /sec. and cofficient of volume
compresebility T� = 8.6 ∗ 10~� ��'�" . Determine the ultimate value of the
compression of the stratum by assuming a uniformly distributed load of 5250 ce HO5⁄ and also determine the time required for 20 percent and 80 percent
consolidation.
Solution : Total compression
�� =T��∆F = 8.6 ∗ 10~� ∗ 10 ∗ 12 ∗ 5250 ∗ ���� = 3.763 in.
For determining the relationship between U% and T for 20%
consolidation use the equation
® = -4¯%100 a2® = 3.144 ∗ 205100 = 0.0314
For 80% consolidation use the equation
® = 1.781 − 0.933 log(100 − 80) Therefore ® = 1.781 − 0.933 log(100 − 80) =0.567
Example
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Example : The loading period for a new building extended from may 1995
to may 1997. In May 2000, the average measured settlement was found to be 11.43
cm. it is known that the ultimate settlement will be about 35.56 cm. Estimate the
settlement in may 2005. Assume double drainage to occur.
Solution:
For the majority of practical cases in which loading is applied over a
period, acceptable accuracy is obtained when calculating time-settlement
relationships by assuming the time datum to be midway through the loading or
construction period.
b� = 11.43`TdℎNMO = 4¬NG2bGMY� = 35.56`T
The settlement is requried for t=9 years, that is up to may 2005. Assuming as
a starting point that at t = 9 years, the degree of consolidation will be = 0.60. under
these conditions. ¯ = 1.13√®.
Is ��� = �NOOcNTNMOGOO�TNO� , ��5 = Settlement at time O5
�����5 = �5 = �®�®5 = �O�O5 ��M`N® = {�O���5
Where �g�±¡' is a constant. Therefore
��.�\¶·' = ¸�ª a2b�5 = 17.15`T
Therefore at t=9 years , ¯ = �¨7\7.7§ = 0.48 and since the value of U is less
than 0.60 the assumption is valid. So the Settlement is 17.15 cm.
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Example :
An oedometer test is performed on a 2 cm thick clay sample. After 5 minutes
50% consolidation is reached. Aftrer how long a time would the same degree of
consolidation be achieved in the field where the clay layer is 3.70 m thick? Assume
the sample and the clay layer have the same drainage boundary conditions (double
drainage).
Solution :
The time factor T is defined as ® = �g��±¡'
Where ��� = half the thickness of the clay for double drainage.
Here , the time factore T and coefficient of consolidation are the same for
both the sample and the field clay layer. The parameter that changes is the time t.
Let O� and O5 be the times required to reach 50% consolidation both in the
oedometer and field respectively. O� = 5T�M. {�O�����5 = {�O5���55
O5 = (�±¡'�±¡})5O� = \¨5
5 ∗ 5 ∗ �§ ∗ �5� YG¬b ≈ 119YG¬b
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Example
Example :
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Example :
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Example:
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Chapter Eight
Shear Strength
•The strength of a material is the greatest stress it can sustain;
•So that the unit of strength is the same as stress (Pa in SI unit system).
Shear Failure in Soils:
Significant of Shear Strength:
• The safety of any geotechnical structure is dependent on the strength of the
soil;
• If the soil fails, the structure founded on it can collapse.
• Understanding shear strength is the basis to analyze soil stability problems
like lateral pressure on earth retaining structures, slope stability, and –
bearing capacity.
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Slope Failure in soils:
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West side of foundation sank 24 ft.
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Shear Strength in Soils •The shear strength of a soil is its resistance to shearing stresses.
•It is a measure of the soil resistance to deformation by continuous displacement of
its individual soil particles
•Shear strength in soils depends primarily on interactions between particles
•Shear failure occurs when the stresses between the particles are such that they
slide or roll past each other
Shear Strength in Soils: Soil derives its shear strength from two sources:
–Cohesion between particles (stress independent component)
• Cementation between sand grains
• Electrostatic attraction between clay particles
–Frictional resistance between particles (stress dependent component)
Shear strength of soils:
Cohesion (C): is a measure of the forces that cement particles of soils.
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Internal Friction angle (φ), is the measure of the shear strength of soils due to
friction
Mohr-Coulomb Failure Criteria
•This theory states that a material fails because of a critical combination of normal
stress and shear stress, and not from their either maximum normal or shear stress
alone.
Mohr Columb Failure criteria
k = ` +�� tan»
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k = ` + �� tan »
Where k : is the shear strength of soil (kPa)
C: cohesion (kPa)
»: is the angle of internal friction (degree)
` :is the cohesion in term of effective stress
»: angle of internal friction in term of effective stress
The above equations are the shear strength of soils in terms total stress and in term
of effective stress.
c = 0 for pure sand ` = 0 for normally consolidated clay
»: Also some time called the drained angle of internal friction.
Mohr-Columb shear Failure criteria
��M» = �� − �\2{`aO»� + �� + �\2
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st = sx t + ½¾¿Àt − ¿ÁvÀ + ½ ½¾¿Àt − ¿ÁvÀ
st = sxÃuv ¤ÄÅ +ÀÂ¥ + ½ Ãuv(ÄÅ + ÀÂ) Also in term of total stress
st =sxÃuv 1ÄÅ +ÀÂ4 + ½Ãuv(ÄÅ + ÀÂ)
Determination of Shear strength Parameters (C , Æ� , À and À)
The shear strength parameters of a soil are determined in the lab primarily with two
types of tests: 1) Direct Shear Test; and 2) Triaxial Shear Test.
-Direct Shear Test:
• The test equipment consists of a metal box in which the soil specimen is placed
• The box is split horizontally into two halves
• Vertical force (normal stress) is applied through a metal platen
• Shear force is applied by moving one half of the box relative to the other to cause
failure in the soil specimen.
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Direct Shear Test Procedure:
1- Measure inner side of shear box and find the area(6 *6 ).
2- Make sure top and bottom halves of shear box are in contact and fixed
together.
3- Place the soil (sand) in three layers in the mold using funnel.
4- Place cover on top of sand.
5- Place the box in machine.
6- Apply normal force. (usually the student divided in to three groups ,
each group use run with one weight, 2 kg, 4 kg and 6 kg the load is
applied through a level arm : Note lever arm loading ratio 1:10 , 2kg
weight =20 kg).
7- Start the motor with selected speed (0.1 in/min) so that the rate of
sharing is at a selected constant rate.
8- Take the horizontal dial reading, vertical dial reading and proving ring
reading (shear load). Record the readings on the data sheet.
9- Continue taking readings until the horizontal shear load peaks and then
falls, or the dial reading gage stopped or at 10% of the the length of the
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sample.
10- Do the calculation :
Find the ¿ÇÈuÉ¿ÃÉÈ¿¿ = ¿ÇÈuÉʾuwËÉÈu(x̽ÍÂ) 11- Plot the results (shear stress Vs horizontal displacement) the
result will be as in figure, take 3 peak stresses. Then plot the relation
between shear strength and normal stresses. Find the slop of this curve,
find , the intersection of the line with y-axis gives the value of C.
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The disadvantages of direct shear test are:
Advantages:
1- The test's simplicity and , in the case of sands, the ease of specimen
preparation
2- The travel of the machine can be reversed to determine the residual shear
strength values, which is shear strength parameters at large displacements.
3- Shear box represents a cheaper method in determining the drained shear
strength parameters for coarse-grained soil.
Disadvantages
1) The main one: drainage conditions cannot be controlled.
2) As pore water pressure cannot be measured, only the total normal stress
can be determined, although this is equal to the effective normal stress if
the pore water pressure is zero.
3) Only an approximation to the state of pure shear is produced in the
specimen
and shear stress on the failure plane is not uniform, failure occurring
progressively from the edges towards the center of the specimen.
4) The area under the shear and vertical loads does not remain constant
throughout the test.
Triaxial shear test :
This experiment over come all the disadvantages of direct shear test:
1- Can be used for all types of soils
2- Pore water pressure can be measured
3- The area corrected be used
4- The failure plane
Triaxial shear test is the most widely used and is suitable for all types of soils. A cylindrical
specimen, generally having a length to diameter ratio of 2, is used in the test and is stressed under
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conditions of axial symmetry in the manner shown in figure.
Soil Sample in triaxial apparatus
A sample of soil in a sealed triaxial test cell. Confining pressure is applied using
water surrounding sample as shown in following figure:
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Figure ( ) triaxial compression testing device
1- Triaxial test is more reliable because we can measure both drained and
undrained shear strength.
2- Generally 1.4” diameter (3” tall) or 2.8” diameter (6” tall) specimen is used.
3- Specimen is encased by a thin rubber membrane and set into a plastic
cylindrical chamber.
4- Cell pressure is applied in the chamber (which represents σ3 by pressurizing
the cell fluid (generally water).
5- Vertical stress is increased by loading the specimen (by raising the platen in
Valve A
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strain controlled test and by adding loads directly in stress controlled test,
but strain controlled test is more common) until shear failure occurs. Total
vertical stress, which is σ1 is equal to the sum of σ3’ and deviator stress
(σd). Measurement of σd, axial deformation, pore pressure, and sample
volume change are recorded.
Depending on the nature of loading and drainage condition, triaxial tests
are conducted in three different ways.
UU Triaxial test (unconsolidated undrained triaxial test)
CU triaxial test (consolidated undrained triaxial test)
CD triaxial test ( consolidated drained triaxial test)
Unconsolidated Undrained Triaxial test:
As drainage is not permitted and consolidation is not necessary, this test is very
quick and also called a quick test. In this test as drainage is not permitted the pore
water pressure increases after application of �\ as well as after the application of �� so ∆ Y NOa�\ = V∆�\ and∆ Y NOa�� = �∆�� so the total excess pore
water pressure will be as follows: ∆ = V∆�\ + �∆��
But �� =�� −�\
This test is common in clayey soils.
Application: UU triaxial test gives shear strength of soil at different confining
stresses. Shear strength is important in all types of geotechnical designs and
analyses.
Test procedure:
1- Measure diameter, length, and initial mass of the specimen.
2- Set a soil specimen in a triaxial cell.
3- Increase the cell pressure to a desired value (100 kPa for the first case and
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200 kPa in the second case and 300 kPa in the third case).
4- Shear the specimen by strain control record b�, and record the ��
5- Continue the test until the proving ring stop reading or use 10%
After completion of test measure of the soil specimen again and put it into the
oven then find the moisture content of the sample.
Take the sample out the cell and Sketch the mode of failure.
Repeat the test for the second specimen too (200 kPa of cell pressure and third
specimen 300 kPa of cell pressure).
calculation :
1- Î� = ∆UU (∆W = �N2O�`GcYNHa2TGO�aMaHOℎNb)N`�TNM).
2- Calculate the corrected area (��): �� = <��~Ï
3- Calculate the stress � = U���<�
4- Plot �� versus axial strain separately for three tests.
5- Plot �� versus � for three tests in the same plot
6- Plot Mohr circle using �� and �\ at failure (note they should give the same �� value)
7- Plot the tangent to the three Mohr circles you plotted in above stage
8- Make a straight line, which is tangent to all three Mohr's circles. The slop give the angle
of internal friction » =0 , the point of the intersection of the tangent and y-axis give the
cohesion (` ) ` = ��2
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Cu triaxial test : consolidated undrained test
In this test consolidation is permitted by opening drainage valve through applying �\,or cell pressure, or all around pressure. Then the drainage valve closed during
applying the deviator stress �� loading.
∆ = V∆�\ + �∆��
Because valve A open during applying cell pressure, therefore B=0
So the increase in pore water pressure is due to increase in deviator stress only.
∆ = �∆��
CD triaxial test : consolidated drained triaxial test
In this test valve A is open all time and the pore water pressure parameters
A and B = 0.
So this test is called drained test also called long term test and slow test.
There is volumetric change:
So �� =� �~Ïg�~ÏÐ whereÎ�: is the volumetric strain Î� = ∆���
In this case: ` = ` and » = » usually occurs in sandy soils
The drained and undrained condition:
Drained condition
• occurs when there is no change in pore water pressure due to external
loading.
• In a drained condition, the pore water can drain out of the soil easily, causing
volumetric strains in the soil.
Undrained condition
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• Occurs when the pore water is unable to drain out of the soil.
• In an undrained condition, the rate of loading is much quicker than the rate
at which the pore water is able to drain out of the soil.
• As a result, most of the external loading is taken by the pore water, resulting
in an increase in the pore water pressure.
• The tendency of soil to change volume is suppressed during undrained
loading.
The existence of either a drained or an undrained condition in a soil
depends on:
• The soil type (e.g. fine-grained or coarse-grained)
• Geological formation (fissures, sand layers in clays, etc.)
• Rate of loading
• For a rate of loading associated with a normal construction activity,
saturated coarse-grained soils (e.g. sands and gravels) experience
drained conditions and saturated fine-grained soils (e.g. silts and clays)
experience undrained conditions
• If the rate of loading is fast enough (e.g. during an earthquake),
even coarse-grained soils can experience undrained loading, often
resulting
in liquefaction.
Unconfined compression test: This test subjects the soil to an axial
compressive load between two platens as shown in the Figure. There is no
confinement of the sample in the radial direction.
The load is recorded using a proving ring or a load cell and the axial deformation of
the soil sample is recorded using a dial gauge.
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Unconfined compression test�\ = 0, only suitable for cohesive soil.
To find the stress
� = F��
�� =�( 11 − Î�) Significance of unconfined compression test
• A quick test to obtain the shear strength parameters of cohesive (fine
grained) soils either in undisturbed or remolded state .
• The test is not applicable to cohesion less or coarse grained soils
• The test is strain controlled and when the soil sample is loaded rapidly, the
pore
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pressures (water within the soil) undergo changes that do not have enough
time to dissipate Hence the test is representative of soils in construction sites
where the rate of construction is very fast and the pore waters do not have
enough time to dissipate.
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Chapter Nine
Lateral Earth Pressure
Lateral earth pressure
Is the pressure that soil exerts in the horizontal direction. The lateral earth
pressure is important because it affect the consolidation behavior and strength of
the soil and because it is considered in the design of engineering structures such as
retaining walls, basements, tunnel, deep foundations.
Coefficient of lateral earth pressure, k is defined as the ratio of the horizontal stress
to the vertical stress.
There are three coefficients:
1- Coefficient of earth
pressure at rest
() = 1 − b�M»
2- Coefficient of active earth
pressure (�) = �~���Ñ�|ÒÓÔÑ
3- Coefficient of passive
earth pressure (*) =�|ÒÓÔÑ�~���Ñ = �]Ð
: When the wall is at rest and the material is in its natural stable then the
pressure applied by material is known as earth pressure at rest
� : when the wall moves away from the backfill , there is a decrease in the
pressure on the wall this decrement continues until a minimum value is reach after
which there is no reduction in the pressure and the value become constant this kind
of pressure is known as active earth pressure.
* : when the wall moves toward the backfill, there is an increase in the pressure
on the wall and this increase continues until maximum value is reach after which
there is no increase in the pressure and the value will become constant ,this kind of
pressure is known as passive earth pressure.
Active
pressure
Soil Mechanics Lectures /Coarse 2-----------------------------2016-2017-------------------------------------------Third year Student
79
Consider the wall shown in the figure
1- If no movement of the wall takes place, the soil is at rest and the vertical and horizontal
stresses acting on element A and B are: